the period of the moon around the earth is 28 days and the radius of i

1.	the period of the moon around the earth is 28 days and the radius of its orbit is 410 to the power of 5 km, if G =6.6710 to the power of -11.Find the mass of the earth?
Answer» The period of the moon is calculated from this equation: $T=\frac{2\pi <klux>R</klux>}{v}.$ Where r is the radius of the orbit and v is the velocity. Since the moon moves in a circular motion, therefore, the centripetal force is: $F_c=\frac{mv^2}{r}$ . In this EXAMPLE, the gravitational force of the Earth acts as the centripetal force. therefore, $F_g=F_c$ . Therefore, $\frac{Gm_mm_e}{r^2}=\frac{m_mv^2}{r}$ . Where $m_m$ is the mass of the moon and $m_e$ is the mass of earth. $\frac{Gm_mm_e}{r^2}=\frac{m_mv^2}{r}$ $\frac{Gm_e}{r}=v^2\\ v=\sqrt{\frac{Gm_e}{r}}$ . Substitute by $v$ in the period equation: $T=\frac{2\pi r}{\sqrt{\frac{Gm_e}{r} } }\\T=2\pi \sqrt{\frac{r^3}{Gm_e} }$ . The required is $m_e$ . But first, we NEED to convert the period from DAYS to seconds. *Therefore, T = 28 days24 hours* 60 MINUTES60 seconds = 2419200 seconds.* Therefore, $2419200=2\pi \sqrt{\frac{(410^510^3)^3}{6.6710^{-11}m_e} }\\385027.6=\sqrt{\frac{9.5910^{35}}{m_e}} \\(385027.6)^2=\frac{9.5910^{35}}{m_e}$ $m_e=\frac{9.5910^{35}}{(385027.6)^2}=6.410^{24}\:kg$ Therefore,

the period of the moon around the earth is 28 days and the radius of its orbit is 410 to the power of 5 km, if G =6.6710 to the power of -11.Find the mass of the earth?

Answer»

The period of the moon is calculated from this equation: $T=\frac{2\pi <klux>R</klux>}{v}.$ Where r is the radius of the orbit and v is the velocity.

Since the moon moves in a circular motion, therefore, the centripetal force is: $F_c=\frac{mv^2}{r}$ .

In this EXAMPLE, the gravitational force of the Earth acts as the centripetal force. therefore, $F_g=F_c$ .

Therefore, $\frac{G*m_mm_e}{r^2}=\frac{m_mv^2}{r}$ . Where $m_m$ is the mass of the moon and $m_e$ is the mass of earth.

$\frac{G*m_mm_e}{r^2}=\frac{m_mv^2}{r}$

$\frac{Gm_e}{r}=v^2\\ v=\sqrt{\frac{Gm_e}{r}}$ .

Substitute by $v$ in the period equation:

$T=\frac{2\pi r}{\sqrt{\frac{Gm_e}{r} } }\\T=2\pi \sqrt{\frac{r^3}{Gm_e} }$ .

The required is $m_e$ . But first, we NEED to convert the period from DAYS to seconds.

Therefore, T = 28 days*24 hours* 60 MINUTES*60 seconds = 2419200 seconds.

Therefore, $2419200=2\pi \sqrt{\frac{(4*10^5*10^3)^3}{6.67*10^{-11}*m_e} }\\385027.6=\sqrt{\frac{9.59*10^{35}}{m_e}} \\(385027.6)^2=\frac{9.59*10^{35}}{m_e}$

$m_e=\frac{9.59*10^{35}}{(385027.6)^2}=6.4*10^{24}\:kg$ Therefore,

the period of the moon around the earth is 28 days and the radius of its orbit is 410 to the power of 5 km, if G =6.6710 to the power of -11.Find the mass of the earth?

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the period of the moon around the earth is 28 days and the radius of its orbit is 4*10 to the power of 5 km, if G =6.67*10 to the power of -11.Find the mass of the earth?

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the period of the moon around the earth is 28 days and the radius of its orbit is 410 to the power of 5 km, if G =6.6710 to the power of -11.Find the mass of the earth?